Solution

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\begin{document}

$14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a
circular orbit of radius $R$?

We can solve this problem using Newton's second law for the radial direction of the motion.
In this case the acceleration is:

\begin{equation} a_r = \frac{v^2}{R} \end{equation}

The force acting on the electron is the electric force:

\begin{equation} F_e = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} \end{equation}

This force points in the radial direction, like the acceleration of the electron.

Using Newton's second law:

\begin{equation} F_e = m_e a_r \rightarrow \frac{m_e v^2}{R} = \frac{(ez)e}{4 \pi \varepsilon_0 R^2} \end{equation}

Then:

\begin{equation} v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}} \end{equation}

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